26. Example 1 In a Stop-and-Wait ARQ system, the bandwidth of the line is 1 Mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth-delay product? If the system data frames are 1000 bits in length, what is the utilization percentage of the link? Solution The bandwidth-delay product is 1 10 6 20 10 -3 = 20,000 bits The system can send 20,000 bits during the time it takes for the data to go from the sender to the receiver and then back again. However, the system sends only 1000 bits. We can say that the link utilization is only 1000/20,000, or 5%. For this reason, for a link with high bandwidth or long delay, use of Stop-and-Wait ARQ wastes the capacity of the link.
27. Example 2 What is the utilization percentage of the link in Example 1 if the link uses Go-Back-N ARQ with a 15-frame sequence? Solution The bandwidth-delay product is still 20,000. The system can send up to 15 frames or 15,000 bits during a round trip. This means the utilization is 15,000/20,000, or 75 percent. Of course, if there are damaged frames, the utilization percentage is much less because frames have to be resent.
28. 11.5 HDLC Configurations and Transfer Modes Frames Frame Format Examples Data Transparency
36. Table 11.1 U-frame control command and response Request information mode RIM Set initialization mode SIM Reset RSET Exchange ID XID Frame reject FRMR Unnumbered information UI Unnumbered acknowledgment UA Request disconnect RD Disconnect DISC Disconnect mode DM Unnumbered poll UP Set asynchronous balanced mode (extended) SABME Set asynchronous balanced mode SABM Set normal response mode (extended) SNRME Set normal response mode Meaning SNRM Command/response
37. Example 3 Figure 11.22 shows an exchange using piggybacking where is no error. Station A begins the exchange of information with an I-frame numbered 0 followed by another I-frame numbered 1. Station B piggybacks its acknowledgment of both frames onto an I-frame of its own. Station B’s first I-frame is also numbered 0 [N(S) field] and contains a 2 in its N(R) field, acknowledging the receipt of A’s frames 1 and 0 and indicating that it expects frame 2 to arrive next. Station B transmits its second and third I-frames (numbered 1 and 2) before accepting further frames from station A. Its N(R) information, therefore, has not changed: B frames 1 and 2 indicate that station B is still expecting A frame 2 to arrive next.
39. Example 4 In Example 3, suppose frame 1 sent from station B to station A has an error. Station A informs station B to resend frames 1 and 2 (the system is using the Go-Back-N mechanism). Station A sends a reject supervisory frame to announce the error in frame 1. Figure 11.23 shows the exchange.
41. Bit stuffing is the process of adding one extra 0 whenever there are five consecutive 1s in the data so that the receiver does not mistake the data for a flag. Note :
